Đặt B, ta có:
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
Thấy:
\(-\frac{1}{2.3}+\frac{1}{2.3}=0;-\frac{1}{3.4}+\frac{1}{3.4}=0\)
\(\Rightarrow2B=\frac{1}{2}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{9900}=\frac{4950}{9900}-\frac{1}{9900}=\frac{4949}{9900}\)
\(\Rightarrow B=\frac{4949}{9900}:2=\frac{4949}{19800}\)